Ihave some questions regarding combinations in finite. here is my first one i am stumped on:
select 3 coins, but assume that there are 5 dimes, 3 nickels, and 2 quarters.
in how many possible ways can the selection be made so that the value of the coins is at least 25
any ? formulas have not done me so well

We will get the number of possible selections, and then subtract the number less than 25 cents.

We can choose the number of dimes 5 ways 0,1,2,3 or 4.
We can choose the number of nickels 4 ways 0,1,2 or 3.
We can choose the number of quarters 3 ways 0,1, or 2.

That's 5*4*3  = 60 selections 

Now we must subtract from the 60 the number of selections of coins that are less than 25 cents. These will involve only dimes and nickels. 

To get a selection of coin worth less than 25 cents:
If we use no dimes, we can use 0,1,2  on all 3 nickels.
That's 4 selections less than 25 cents. (that includes the choice of No coins at all in the 60, which we must subtract).

If we use exactly 1 dime , we can use 0,1,2, or all 3 nickels.
That's the 3 combinations less than 25 cents.

And there is 1 other selection less than 25 cents, 2 dimes and no nickels. 

So that's 4+3+1 = 8 selections which we must subtract from the 60.
 
Answer 60-8 = 52 selections of coins worth 25 cents or more.

Just a second zzz